問題描述:例如:當前路徑:127.0.0.1:8080/test/content/content.action 返回到content頁面
問題:如何在content.jsp頁面如何獲取:127.0.0.1:8080/test/content/content.action ?
注:String url=request.getScheme()+"://" + request.getHeader("host") + request.getRequestURI()+request.getQueryString();
這個獲取的是:127.0.0.1:8080/test/web-inf/view/content.jsp?
如何獲取:127.0.0.1:8080/test/content/content.action 路徑?
js ------> thisURL=document.URL